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I am trying to solve the following integral using the steepest descent method:

$$I_n(\alpha)=\int_0^1x^{\alpha n} (1-x)^{\beta n} dx$$

$\beta=1-\alpha , \alpha > 0, \beta > 0$.

This is from a book, and the solution says that the saddle point is at $x=\alpha$. I cannot see why. I tried to put it in the form: $$\int_a^be^\frac{-f(x)}{\lambda}dx$$

To do this, I used $x^{\alpha n}=e^{\alpha n \ln(x)}$ and $(1-x)^{\beta n} = e^{\beta n\ln(1-x)}$. After simplifying, this gave me $\lambda=\frac{1}{n}$ and $f(x)=\ln(x^\alpha(1-x)^{1-\alpha})$. To minimise $f(x)$, I don't see how $x$ can equal $\alpha$. Unfortunately, the book doesn't give more information on how the saddle point is at $x=\alpha$. I would like to know how to obtain this result and complete the integral.

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  • $\begingroup$ Comment to the post (v1): Presumably $n$ is assumed large. $\endgroup$ – Qmechanic Nov 17 '17 at 15:22
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For any $n\in\mathbb{N}$ and $\alpha\in(0,1)$ we have:

$$\begin{eqnarray*} I_n(\alpha)=\int_{0}^{1}\left[x^\alpha(1-x)^{1-\alpha}\right]^n\,dx &=& B(n\alpha+1,n-n\alpha+1)\\&=&\frac{\Gamma(n\alpha+1)\,\Gamma(n-n\alpha+1)}{\Gamma(n+1)}\end{eqnarray*} $$ by Euler's Beta function. Asymptotics can be derived from Gautschi's inequality and/or Stirling's approximation. As an alternative, the maximum value of $f(x)=x^\alpha(1-x)^{1-\alpha}$ over $[0,1]$ is attained at $x=\alpha$, and $f''(\alpha)=-\frac{1}{\alpha}\left(\frac{\alpha}{1-\alpha}\right)^\alpha$. It follows that

$$ I_n(\alpha)\approx \alpha^{n\alpha}(1-\alpha)^{n-n\alpha}\int_{-\infty}^{+\infty}e^{-\frac{n}{2\alpha}\left(\frac{\alpha}{1-\alpha}\right)^\alpha z^2}\,dz=\alpha^{n\alpha}(1-\alpha)^{n-n\alpha}\sqrt{\frac{2\pi}{\frac{n}{\alpha}\left(\frac{\alpha}{1-\alpha}\right)^\alpha}}. $$

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  • $\begingroup$ Thanks. That was really insightful. $\endgroup$ – Adil Patel Aug 20 '17 at 19:10
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\begin{align} \frac {\mathrm d}{\mathrm d x}(x^\alpha(1-x)^{1-\alpha}) &= ((1-x)\alpha - x(1-\alpha))(x^{\alpha-1}(1-x)^{-\alpha}) \\ &= (\alpha-x)(x^{\alpha-1}(1-x)^{-\alpha}) \end{align}

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